SOLUTION: Solve the simultaneous equations : logbase2(x) + logbase8(y) = -1 logbase4(x) + logbase2(y) = 2 Cannot solve, spent hours on it Any help would be appreciated

Algebra ->  Algebra  -> Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Solve the simultaneous equations : logbase2(x) + logbase8(y) = -1 logbase4(x) + logbase2(y) = 2 Cannot solve, spent hours on it Any help would be appreciated      Log On

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 Question 565979: Solve the simultaneous equations : logbase2(x) + logbase8(y) = -1 logbase4(x) + logbase2(y) = 2 Cannot solve, spent hours on it Any help would be appreciatedAnswer by lwsshak3(6513)   (Show Source): You can put this solution on YOUR website!Solve the simultaneous equations : logbase2(x) + logbase8(y) = -1 logbase4(x) + logbase2(y) = 2 ** Change to base2: log2(x) + log2(y)/log2(8) = -1 log2(x)/log2(4) + log2(y) = 2 .. log2(4)=2 log2(8)=3 .. log2(x) + log2(y)/3= -1 log2(x)/2 + log2(y) = 2 .. 3log2(x) + log2(y)= -3 log2(x) +2 log2(y) =4 .. 6log2(x) + 2log2(y)= -6 log2(x) +2 log2(y) =4 subtract 6log2(x)-log2(x)=-10 log2(x^6)-log2(x)=-10 log2(x^6/x)=-10 log2(x^5)=-10 convert to exponential form: base(2) raised to log of number(-10)=number(x^5) 2^-10=x^5 take 5th root of both sides x=(2^-2)=1/(2^2)=1/4 .. solving for y log2(x) +2 log2(y) =4 log2(2^-2)+2log2(y)=4 -2+2log2(y)=4 2log2(y)=6 log2(y)=3 y=8 ans: x=1/4 y=8